Answer
a) The capacitance of the capacitor is $=2.65\times10^{-12}F$
b) The voltage or potential difference is $16,400 V$ or $16.4 kV$
c) The electric field is $5.00\times10^{6} V/m$
Work Step by Step
a) The capacitance is given by the formula $C = \frac{ε_{0}A}{d}$.
Therefore the capacitance of the capacitor is
$$\frac{8.854\times10^{-12}F/m \times 9.82\times10^{-4}m^{2}}{0.00328 m}$$. $$=2.65\times10^{-12}F$$
b) The potential that is stored on a capacitor is given by the equation: $V_{ab} = \frac{Q}{C}$.
Therefore the voltage is $\frac{4.35\times10^{-8}C}{2.65\times10^{-12}F}$ $= 16,400 V$ or $16.4 kV$
c) Since the plates are charged there is also a uniform electric field which can be found from this formula: $E=\frac{V}{d}$. Therefore the strength of the electric field is $E = \frac{16400 V}{0.00328 m}$ $=5.00\times10^{6} V/m$